Protein Synthesis

AS · 14 min

A gene is a stretch of DNA whose base sequence codes for the amino acid sequence of a polypeptide. Turning that code into a protein takes two stages: transcription, in which the base sequence of the gene is copied into messenger RNA in the nucleus, and translation, in which ribosomes in the cytoplasm read the mRNA and join amino acids in the right order, brought by transfer RNA. This note covers the genetic code, both stages step by step, and the splicing of introns out of eukaryotic RNA. It is one of the most heavily examined topics in AS Biology: expect long "describe" questions, and short questions that ask you to convert between DNA, mRNA, anticodons and amino acids using a codon table.

Genes and the genetic code

Definition

A gene is a sequence of nucleotides that forms part of a DNA molecule and codes for a polypeptide. A polypeptide is coded for by a gene.

There are 20 amino acids but only 4 bases. One base per amino acid would give only 4 combinations, two bases only 42=164^2 = 16. Three bases give 43=644^3 = 64 combinations, enough for all 20 amino acids. So the code is a triplet code.

Key result

Features of the genetic code

  • Triplet code: a sequence of three bases codes for one amino acid. A triplet on mRNA is called a codon.
  • Universal: the same triplets code for the same amino acids in (almost) all organisms. This is why a human gene inserted into a bacterium produces the human protein.
  • Degenerate: most amino acids are coded for by more than one triplet. Of the 64 codons, 61 code for amino acids and 3 are stop codons.
  • Start and stop codons: the codon AUG (DNA triplet TAC on the template strand) codes for methionine and acts as the start signal for translation. UAA, UAG and UGA are stop codons, which do not code for any amino acid and signal the end of the polypeptide.
  • Non-overlapping: each base is part of only one codon; the code is read in consecutive groups of three from a fixed starting point.

The mRNA codon table (first base on the left, second base across the top, third base within each cell):

First baseUCAG
UUUU Phe
UUC Phe
UUA Leu
UUG Leu
UCU Ser
UCC Ser
UCA Ser
UCG Ser
UAU Tyr
UAC Tyr
UAA stop
UAG stop
UGU Cys
UGC Cys
UGA stop
UGG Trp
CCUU Leu
CUC Leu
CUA Leu
CUG Leu
CCU Pro
CCC Pro
CCA Pro
CCG Pro
CAU His
CAC His
CAA Gln
CAG Gln
CGU Arg
CGC Arg
CGA Arg
CGG Arg
AAUU Ile
AUC Ile
AUA Ile
AUG Met (start)
ACU Thr
ACC Thr
ACA Thr
ACG Thr
AAU Asn
AAC Asn
AAA Lys
AAG Lys
AGU Ser
AGC Ser
AGA Arg
AGG Arg
GGUU Val
GUC Val
GUA Val
GUG Val
GCU Ala
GCC Ala
GCA Ala
GCG Ala
GAU Asp
GAC Asp
GAA Glu
GAG Glu
GGU Gly
GGC Gly
GGA Gly
GGG Gly

You do not need to learn the table; the exam provides whatever part is needed. You must know AUG (start) and that there are three stop codons.

The template and non-transcribed strands

Only one of the two DNA strands of a gene is used to make mRNA.

Definition

The strand of a DNA molecule that is used in transcription is called the transcribed or template strand. The other strand is called the non-transcribed strand.

Because mRNA is complementary to the template strand, it has the same base sequence as the non-transcribed strand, except that it contains U where the DNA has T. (The non-transcribed strand is sometimes called the coding strand for this reason.)

Example sequence
Non-transcribed strand (DNA)5′ ATG GCT TTC CAA TGA 3′
Template strand (DNA)3′ TAC CGA AAG GTT ACT 5′
mRNA (complementary to template)5′ AUG GCU UUC CAA UGA 3′
Amino acidsMet – Ala – Phe – Gln – stop

Transcription

Transcription takes place in the nucleus. DNA is too large to leave the nucleus and must be protected, so a mobile copy of the gene, mRNA, is made instead.

Transcription
  1. The DNA double helix unwinds at the gene, and the hydrogen bonds between bases are broken, separating the two strands over a short section. (In eukaryotes this is done by RNA polymerase together with other proteins.)
  2. One strand, the template strand, is used. Its bases are exposed.
  3. Free RNA nucleotides (activated, carrying extra phosphates) in the nucleus pair with the exposed bases by complementary base pairing: A on the template pairs with U, T with A, C with G and G with C. Hydrogen bonds form between them temporarily.
  4. RNA polymerase moves along the template strand, joining the RNA nucleotides by forming phosphodiester bonds between them, building a single strand of RNA in the 5′ → 3′ direction.
  5. Behind the RNA polymerase, the RNA separates from the template, and the DNA strands rejoin (hydrogen bonds re-form) and rewind.
  6. When RNA polymerase reaches the end of the gene (a terminator sequence), it detaches, and the RNA molecule is released.

In eukaryotes, the RNA produced directly by transcription is called the primary transcript (or pre-mRNA). It is modified before it leaves the nucleus.

Splicing: removing introns

Eukaryotic genes contain sections that do not code for amino acids.

Definition
  • Introns are non-coding sequences within a gene.
  • Exons are the coding sequences of a gene.
  • Splicing: in eukaryotes, the primary transcript is modified by the removal of introns and the joining together of exons to form mRNA.

Both introns and exons are transcribed, so the primary transcript contains both. Splicing (carried out by complexes of protein and RNA) cuts out the introns and joins the exons end to end. The finished mRNA is shorter and contains only the coding sequence, which can be translated continuously. It then leaves the nucleus through a nuclear pore and travels to a ribosome in the cytoplasm or on the rough endoplasmic reticulum.

Prokaryotic genes do not have introns, so prokaryotes do not splice their RNA; and because they have no nucleus, translation can begin while transcription is still going on.

Translation

Translation takes place at ribosomes in the cytoplasm (free or on the rough ER). The ribosome reads the mRNA codon by codon, and tRNA molecules bring the matching amino acids.

Transfer RNA

Each tRNA molecule is a single strand of RNA, about 80 nucleotides long, folded into a clover-leaf shape held by hydrogen bonds between complementary bases within the strand. It has:

  • an anticodon: a sequence of three bases at one end (in a loop) that is complementary to a particular mRNA codon;
  • an amino acid binding site at the other end (the 3′ end), where the specific amino acid that corresponds to the anticodon is attached.

Each amino acid is attached to its own tRNA by a specific enzyme, using energy from ATP. This "activates" the amino acid ready for peptide bond formation.

Ribosomes

A ribosome has a small subunit and a large subunit, made of rRNA and protein. It has a groove for the mRNA and room for two tRNA molecules at a time, so it holds two codons (six bases) of the mRNA in place while the amino acids on the two tRNAs are joined.

Translation
  1. The mRNA binds to the small subunit of a ribosome, at its 5′ end. The ribosome moves to the start codon, AUG.
  2. A tRNA with the complementary anticodon UAC, carrying methionine, binds to the start codon by complementary base pairing (hydrogen bonds between codon and anticodon).
  3. A second tRNA, with an anticodon complementary to the next codon, binds alongside it, bringing the second amino acid.
  4. A peptide bond forms between the two amino acids, by a condensation reaction catalysed within the ribosome (by rRNA in the large subunit). The amino acid chain is now attached to the second tRNA.
  5. The ribosome moves along the mRNA by one codon (three bases). The first tRNA, now without an amino acid, is released and can pick up another molecule of its specific amino acid from the cytoplasm.
  6. A third tRNA binds to the next codon, its amino acid is joined to the chain by a peptide bond, and so on. The polypeptide grows one amino acid at a time.
  7. When the ribosome reaches a stop codon (UAA, UAG or UGA), no tRNA can bind. The completed polypeptide is released and the ribosome subunits separate from the mRNA.

Several ribosomes can translate the same mRNA molecule at once, one behind the other, forming a polysome, so many copies of the polypeptide are made quickly. After translation, polypeptides made on the rough ER are folded and modified (for example, by adding carbohydrate in the Golgi body) to form functional proteins.

Key result

Roles of the molecules in protein synthesis

MoleculeRole
DNA (template strand)provides the base sequence that is copied into mRNA
RNA polymerasejoins RNA nucleotides with phosphodiester bonds during transcription, using the template strand
mRNAcarries a copy of the genetic code of a gene from the nucleus to the ribosome; a sequence of codons
Codontriplet of bases on mRNA that codes for one amino acid (or start/stop)
tRNAcarries a specific amino acid to the ribosome; its anticodon pairs with the complementary codon
Anticodontriplet of bases on tRNA, complementary to a codon, which ensures the correct amino acid is added
Ribosomeholds mRNA and two tRNAs in place; catalyses peptide bond formation; moves along the mRNA one codon at a time
ATPprovides energy for attaching amino acids to tRNA and for movement of the ribosome
DNA to polypeptide (routine)

The template strand of part of a gene has the sequence

3′ TAC CGA AAG GTT ACT 5′

(a) Write the mRNA sequence transcribed from it. (b) Use the codon table to give the amino acid sequence. (c) Give the anticodons of the tRNA molecules that bring the first four amino acids.

Solution

(a) mRNA is complementary to the template (with U instead of T): 5′ AUG GCU UUC CAA UGA 3′.

(b) AUG = Met (start), GCU = Ala, UUC = Phe, CAA = Gln, UGA = stop. Polypeptide: Met–Ala–Phe–Gln.

(c) Anticodons are complementary to the codons: UAC, CGA, AAG, GUU.

Notice that the anticodons have the same sequence as the template DNA triplets, but with U instead of T.

Working backwards (moderate)

A tRNA molecule has the anticodon GAC.

(a) Give the mRNA codon it pairs with, and the amino acid it carries. (b) Give the base sequence of the triplet on (i) the template strand and (ii) the non-transcribed strand of DNA. (c) A mutation changes this template triplet from GAC to GAG. Explain why the polypeptide is unchanged.

Solution

(a) Codon: CUG (complementary to GAC). From the table, CUG = leucine (Leu).

(b) (i) The template strand is complementary to the mRNA codon CUG: GAC. (ii) The non-transcribed strand has the same sequence as the mRNA, with T for U: CTG.

(c) The template triplet GAG is transcribed into the mRNA codon CUC, which also codes for leucine. The genetic code is degenerate: leucine has six codons (UUA, UUG, CUU, CUC, CUA, CUG), so the same amino acid is inserted and the primary structure is unchanged.

(Always convert step by step: template DNA → mRNA (complement, with U) → codon table.)

Introns and exons (moderate)

A eukaryotic gene contains four exons of 150, 210, 96 and 450 nucleotides, separated by three introns of 800, 1200 and 650 nucleotides. Assume the exons contain only the coding sequence, from the start codon to the stop codon.

(a) Calculate the length of the primary transcript and of the mRNA. (b) Calculate the percentage of the primary transcript removed by splicing. (c) Calculate the number of amino acids in the polypeptide.

Solution

(a) Exons: 150+210+96+450=906150 + 210 + 96 + 450 = 906 nucleotides. Introns: 800+1200+650=2650800 + 1200 + 650 = 2650 nucleotides.

Primary transcript =906+2650=3556= 906 + 2650 = 3556 nucleotides. mRNA =906= 906 nucleotides.

(b) 26503556×100=74.5%\dfrac{2650}{3556} \times 100 = 74.5\%.

(c) 906/3=302906 / 3 = 302 codons. The last is a stop codon, which does not code for an amino acid, so the polypeptide has 301 amino acids (including the methionine coded by the start codon).

Exam-hard: describing translation (6 marks)

Describe how a polypeptide is synthesised at a ribosome from a molecule of mRNA.

Solution

Any six of:

  1. mRNA binds to the small subunit of the ribosome;
  2. the ribosome finds the start codon AUG;
  3. tRNA molecules carry specific amino acids (attached using ATP);
  4. a tRNA with a complementary anticodon binds to the codon by complementary base pairing / hydrogen bonds;
  5. the ribosome holds two codons / two tRNAs at once;
  6. a peptide bond forms between adjacent amino acids (condensation), catalysed by the ribosome;
  7. the ribosome moves along the mRNA one codon at a time; the first tRNA is released (to collect another amino acid);
  8. this continues, adding amino acids in the order determined by the sequence of codons;
  9. until a stop codon is reached, and the polypeptide is released.
Watch out
  • In transcription, A on DNA pairs with U on RNA, not T. Forgetting uracil is the commonest error.
  • RNA polymerase makes mRNA. DNA polymerase is for replication only.
  • The codon is on mRNA; the anticodon is on tRNA. Do not write "the DNA codon" or "the mRNA anticodon". (On DNA, call it a base triplet.)
  • Peptide bonds form between amino acids, not between tRNA molecules or between codons and anticodons (those are hydrogen bonds).
  • Introns are removed from the primary transcript (RNA), not from the DNA; the gene still contains its introns.
  • A stop codon does not code for an amino acid, so the number of amino acids is (number of codons − 1) when the stop codon is included.
  • Translation happens at ribosomes, in the cytoplasm or on the RER, not in the nucleus.
Exam tip
  • "Describe transcription" (about 5 marks): DNA unwinds/hydrogen bonds break; template strand; free RNA nucleotides; complementary base pairing (A–U); RNA polymerase joins nucleotides (phosphodiester bonds); mRNA released and leaves nucleus via nuclear pore; in eukaryotes, introns removed / exons joined.
  • "Describe translation" (about 6 marks): see the worked example. Mention anticodon, codon, complementary, peptide bond, ribosome moves, start and stop codons.
  • In conversions, write each step on its own line: template DNA → mRNA → anticodon → amino acid. Check whether you are given the template strand or the non-transcribed strand.
  • When asked about the genetic code, use the four words: triplet, universal, degenerate, non-overlapping, plus start and stop codons.
Summary
  • A gene is a sequence of nucleotides in DNA that codes for a polypeptide.
  • The genetic code is a triplet code; universal; degenerate (61 codons for 20 amino acids, 3 stop codons); non-overlapping; AUG is the start codon (methionine).
  • The template (transcribed) strand is used in transcription; the other is the non-transcribed strand; mRNA has the same sequence as the non-transcribed strand with U for T.
  • Transcription (nucleus): DNA unwinds, RNA nucleotides pair with the template strand (A–U), RNA polymerase forms phosphodiester bonds.
  • In eukaryotes, the primary transcript is spliced: introns removed, exons joined, giving mRNA.
  • Translation (ribosome): tRNA anticodons pair with mRNA codons; ribosome holds two tRNAs; peptide bonds form; ribosome moves one codon at a time; stop codon releases the polypeptide.

Practice questions

Question
  1. Define the term gene.
  2. Explain why the genetic code must be at least a triplet code.
  3. State what is meant by saying the genetic code is (a) universal and (b) degenerate.
  4. Distinguish between the template strand and the non-transcribed strand.
  5. The non-transcribed strand of DNA reads 5′ ATGAAAGGCTAA 3′. Write the mRNA sequence and use the codon table to give the polypeptide.
  6. State two differences between transcription and DNA replication.
  7. Distinguish between a codon and an anticodon.
  8. Explain the role of tRNA in protein synthesis. (3 marks)
  9. A polypeptide contains 450 amino acids. Calculate the minimum number of nucleotides in the mRNA coding sequence, including the stop codon, and explain why the gene in a eukaryotic cell is likely to be much longer.
  10. Describe how the information in a gene is used to produce a polypeptide, starting from DNA in the nucleus. (9 marks)
Answers
  1. A sequence of nucleotides that forms part of a DNA molecule and codes for a polypeptide.
  2. There are 20 amino acids; with 4 bases, a single-base code gives only 4 combinations and a two-base code 42=164^2 = 16, too few; a triplet gives 43=644^3 = 64, enough to code for all 20.
  3. (a) The same codons code for the same amino acids in (almost) all organisms. (b) Most amino acids are coded for by more than one codon/triplet.
  4. The template (transcribed) strand is the DNA strand used as the template for transcription; mRNA is complementary to it. The non-transcribed strand is the other strand of the gene; it is not used, and its sequence is the same as the mRNA (with T in place of U).
  5. mRNA: 5′ AUG AAA GGC UAA 3′. Polypeptide: Met–Lys–Gly (UAA = stop).
  6. Any two: transcription copies only one gene, replication the whole DNA molecule; transcription uses one strand as template, replication both; transcription uses RNA nucleotides (with ribose and uracil), replication DNA nucleotides; transcription uses RNA polymerase, replication DNA polymerase (and helicase, ligase); transcription produces single-stranded RNA, replication two double-stranded DNA molecules; transcription occurs throughout interphase (when the gene is needed), replication only in S phase.
  7. A codon is a triplet of bases on mRNA that codes for an amino acid (or start/stop); an anticodon is a triplet of bases on tRNA that is complementary to a codon.
  8. Each tRNA carries a specific amino acid (attached at one end, using ATP); its anticodon binds to the complementary codon on the mRNA at the ribosome by complementary base pairing; so the amino acids are brought to the ribosome and placed in the order specified by the sequence of codons; tRNA is released after its amino acid has been joined to the chain, to be reused.
  9. 450×3=1350450 \times 3 = 1350 nucleotides, plus 3 for the stop codon =1353= 1353. The gene contains introns (non-coding sequences) between the exons, which are transcribed but removed by splicing; it also has regulatory and non-translated sequences.
  10. Any nine: DNA unwinds and hydrogen bonds between bases break (in the region of the gene); one strand is the template strand; free RNA nucleotides pair with complementary bases (A–U, T–A, C–G, G–C); RNA polymerase joins RNA nucleotides with phosphodiester bonds; the primary transcript is spliced: introns removed and exons joined to form mRNA; mRNA leaves the nucleus through a nuclear pore; mRNA attaches to a ribosome (small subunit); translation begins at the start codon AUG; tRNA molecules carry specific amino acids; the anticodon on tRNA pairs with the complementary codon on mRNA; the ribosome holds two tRNAs/two codons; a peptide bond forms between the amino acids (condensation); the ribosome moves along one codon and the first tRNA is released; process repeats until a stop codon is reached and the polypeptide is released; the order of amino acids (primary structure) is determined by the sequence of codons, i.e. by the base sequence of the gene.

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